User-Defined Parameters

Parameter Value Variable
Series Code 21864 ts
Maximum Lag 48 af.lags
Prevision Horizon 12 n.ahead
Unit Root Test ADF, drift, 6, BIC, 5pct ur.test
ARCH Test 12, FALSE, 0.05 arch.test
Dummy FALSE dummy

Getting the Time Series from the BETS database

library(BETS)
data = BETS.get(code)

Information About the Series

info <- BETS.search(code = ts, view = F)
print(params$teste)
## NULL
Code Description Periodicity Start Source Unit NA
21864 Physical Production - Intermediate goods Index 2002.1 01/01/2002 mar/2017 IBGE

Graph

## library(mFilter)
trend = fitted(hpfilter(data))

library(dygraphs)
dygraph(cbind(Series = data, Trend = trend), main = info[,"Description"]) %>%
  dyRangeSelector(strokeColor = "gray", fillColor = "gray") %>%
    dyAxis("y", label = info[,"Unit"])

Unit Root Tests

Augmented Dickey-Fuller

  test.params = append(list(y = data), ur.test)
  df = do.call(BETS.ur_test,test.params)
  df$results
##      statistic crit.val rej.H0
## tau2 -4.238853    -2.88    yes
## phi1  8.984243     4.63     no

For a 95% confidence interval, the test statistic tau3 is smaller than the critical value. We therefore conclude that there is no non-seasonal unit root.

ns_roots = 0
d_ts = data

Osborn-Chui-Smith-Birchenhall

This test will be performed for lag 12, that is, the frequency of the series 21864.

library(forecast)
s_roots = nsdiffs(data)
print(s_roots)
## [1] 0

According to the OCSB test, there is no seasonal unit root, at least at a 5% significance level.

Auto-Correlation Functions

ACF and PACF - Original Series

BETS.corrgram(d_ts, lag.max = af.lags, mode = "bartlett", knit = T)
BETS.corrgram(d_ts, lag.max = af.lags, mode = "simple", type = "partial", knit = T)

Model Identification and Estimation

The correlograms from last section gives us enough information to try to identify the underlying SARIMA model parameters. We can confirm our guess by running the auto.arima function from the package forecast. By default, this function uses the AICc (Akaike Information Criterion with Finite Sample Correction) for model selection. Here, we are going to use BIC (Bayesian Information Criterion), in which the penalty term for the number of parameters in the model is larger than in AIC.

model <- auto.arima(data, ic = tolower(inf.crit), test = tolower(ur.test$mode), 
                   max.d = ns_roots, max.D = s_roots)
summary(model)
## Series: data 
## ARIMA(3,0,1)(1,0,0)[12] with non-zero mean 
## 
## Coefficients:
##          ar1      ar2      ar3      ma1    sar1     mean
##       1.3701  -0.1013  -0.3325  -0.6919  0.8481  90.8723
## s.e.  0.1431   0.1668   0.0698   0.1430  0.0360   4.7822
## 
## sigma^2 estimated as 7.37:  log likelihood=-454.74
## AIC=923.47   AICc=924.1   BIC=946.05
## 
## Training set error measures:
##                     ME     RMSE      MAE       MPE     MAPE      MASE
## Training set 0.1649182 2.670634 2.037619 0.1003209 2.178236 0.4665688
##                     ACF1
## Training set -0.02589043

We see that, according to BIC, the best model is a SARIMA(3,0,1)(1,0,0)[12]. Nevertheless, this is not the end. We still have to test for heteroskedasticity in the residuals. We can use an ARCH test with this purpose.

arch.params <- append(list(x = resid(model)), arch.test)
at <- do.call(BETS.arch_test, arch.params)
at
##   statistic    p.value   htk
## 1  25.26719 0.01360626 FALSE

The p.value of 0.01 is smaller than the significance level of 0.05. We therefore conclude that the residuals are not heteroskedastic.

The next function outputs the model’s standardized residuals. If they are all inside the confidence interval, it means the behaviour of the series was well captured by the model.

rsd <- BETS.std_resid(model, alpha = 0.01)

Forecasts

BETS.predict(model,h=n.ahead, main = info[,"Description"], ylab = info[,"Unit"], knit = T)